Implications (2) \(\Rightarrow \)  (1) and (2) \(\Rightarrow \)  (3) are straightforward consequences of Theorem REF .
(1) \(\Rightarrow \)  (2): Note that the topology \(\tau _{k}\)  on \(C(X)\)  is \(\tau _{p}\) -polar, i.e. \(\tau _{k}\)  admits a base of absolutely convex neighbourhoods of zero consisting of \(\tau _{p}\) -closed sets. Hence, if \(S\)  is a \(\tau _{k}\) -bounded set in \(C_{k}(X)\) , then
the closure \(\mathop {\it cl}_{\tau _{p}}(S)\)  is still \(\tau _{k}\) -bounded.
Assume that (1) holds. Since \(\tau _{p}\le \tau _{k}\) , the strong topology \(\beta _{X}\)  of \(C_{p}(X)^{\prime }\)  is finer than the strong topology
\(\vartheta _{X}\)  of \(C_{k}(X)^{\prime }\)  restricted to \(C_{p}(X)^{\prime }\)  (shortly \(\vartheta _{X}|C_{p}(X)^{\prime }\le \beta _{X}\) ). As the topology \(\vartheta _{X}\)  is the finest locally convex topology on \(C_{k}(X)^{\prime }\) , we have in fact the equality \(\vartheta _{X}|C_{p}(X)^{\prime }=\beta _{X}\) . Hence \((C_{p}(X)^{\prime },\beta _{X})\)  is a topological subspace of
\((C_{k}(X)^{\prime },\vartheta _{X})\)  and carries the finest locally convex topology, so \(C_{p}(X)\)  is distinguished by Theorem REF .
Next we show that \(\tau _{p}=\tau _{k}\) .
We know that for every absolutely convex neighbourhood of zero \(U\)  in the space \((C_{p}(X)^{\prime },\beta _{X})\)  there exists an absolutely convex neighbourhood of
zero \(V\)  in \((C_{k}(X)^{\prime },\vartheta _{X})\)  such that \(V\cap C_{p}(X)^{\prime }\subset U\) .
Consequently, for each \(\tau _{p}\) -closed bounded absolutely convex set \(B\subset C_{p}(X)\)  there exists an absolutely convex \(\tau _{k}\) -bounded set
\(S\subset C_{k}(X)\)  with \(S^{\bullet }\cap C_{p}(X)^{\prime }\subset B^{\circ }\) , where
\(S^{\bullet }=\lbrace x^*\in C_{k}(X)^{\prime }: |x^*(f)|\le 1, f\in S\rbrace ,\) 
\(B^{\circ }=\lbrace x^*\in C_{p}(X)^{\prime }:|x^*(f)|\le 1, f\in B\rbrace .\) 
Hence, \(\lbrace x^*\in C_{p}(X)^{\prime }:|x^*(f)|\le 1, f\in \mathop {\it cl}_{\tau _{p}}(S)\rbrace \subset \lbrace x^*\in C_{p}(X)^{\prime }:|x^*(f)|\le 1, f\in S\rbrace \subset \lbrace x^*\in C_{p}(X)^{\prime }:|x^*(f)|\le 1, f\in B\rbrace ,\) 
what shows that the polar set \(B^{\circ \circ }= \lbrace x^*\in C_{p}(X)^{\prime }:|x^*(f)|\le 1, f\in B\rbrace ^{\circ }\)  is contained in the polar set
\((\mathop {\it cl}_{\tau _{p}}(S))^{\circ \circ }= \lbrace x^*\in C_{p}(X)^{\prime }:|x^*(f)|\le 1, f\in \mathop {\it cl}_{\tau _{p}}(S)\rbrace ^{\circ }\) , where the corresponding polars are taken in the
dual pair \((C_{p}(X)^{\prime }, C_{p}(X))\) .
By the bipolar theorem [1]}, \(B = B^{\circ \circ }\subset (\mathop {\it cl}_{\tau _{p}}(S))^{\circ \circ }=\mathop {\it cl}_{\tau _{p}}(S)\) .
Since \(\mathop {\it cl}_{\tau _{p}}(S)\)  is \(\tau _{k}\) -bounded,
every \(\tau _{p}\) -bounded set in \(C(X)\)  is \(\tau _{k}\) -bounded.
On the other hand, \(C_{p}(X)\)  is quasibarrelled, i.e. every absolutely convex closed set absorbing bounded sets in \(C_{p}(X)\)  is a neighbourhood of zero, see [1]}.
Now take a \(\tau _{p}\) -closed absolutely convex neighbourhood of zero \(U\)  in \(\tau _{k}\) . Since, as we proved, every \(\tau _{p}\) -bounded set is \(\tau _{k}\) -bounded, the set \(U\)  absorbs bounded sets
in \(C_{p}(X)\) . Hence, \(U\)  is also a neighbourhood of zero in \(\tau _{p}\) , so \(\tau _{p}=\tau _{k}\) .
(3) \(\Rightarrow \)  (2): Take a bounded set \(A\subset C_{p}(X)\) . Then there exists a bounded set \(B\subset C_{k}(X)\)  with
\(A \subset \mathop {\it cl}_{\mathbb {R}^{X}}(B) \cap C_{p}(X)(\subset \mathop {\it cl}_{\tau _{p}}(B)\) ).
Clearly, the closure \(\mathop {\it cl}_{\tau _{p}}(B)\)  is still \(\tau _{k}\) -bounded. Therefore, the topologies \(\tau _{p}\)  and \(\tau _{k}\)  have the same bounded sets,
and then \(\tau _{p}=\tau _{k}\) . Applying Theorem REF  we deduce that \(C_{p}(X)\)  is distinguished.
